# Two cars left the city for a suburb, 120 km away, at the same time. The speed of one of the cars was 20 km/hour greater than the speed of the other, and that is why it arrived at the suburb 1 hour before the other car. What is the speed of both cars.?

Apr 7, 2017

Slower car: $40 \text{km/h}$
Faster car: $60 \text{km/h}$

#### Explanation:

Let $x =$hours it took the slower car to reach the suburb
The time it took the faster car to reach the suburb is $x - 1$

Since distance$\div$time$=$speed,
The speed of the slower car is $120 \text{km"div(x)"h"=120/x"km/h}$
The speed of the faster car is $120 \text{km"div(x+1)"h"=120/(x-1)"km/h}$

Since the faster car travelled $20 \text{km/h}$ faster than the slower car,
$\frac{120}{x - 1} = \frac{120}{x} + 20$

Now multiply both sides by $\left(x - 1\right)$ and then $x$ to start solving.
$\frac{120 \left(x - 1\right)}{x - 1} = \frac{120 \left(x - 1\right)}{x} + 20 \left(x - 1\right)$
$\frac{120 x \left(x - 1\right)}{x - 1} = \frac{120 x \left(x - 1\right)}{x} + 20 x \left(x - 1\right)$

Simplify, distribute, factor, disregard, etc. Essentially, solve.
$\frac{120 x \textcolor{red}{\cancel{\textcolor{b l a c k}{x - 1}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 1}}}} = \frac{120 \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} \left(x - 1\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} + 20 x \left(x - 1\right)$
$\textcolor{red}{\cancel{\textcolor{b l a c k}{120 x}}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{120 x}}} - 120 + 20 {x}^{2} - 20 x$
$0 = 20 {x}^{2} - 20 x - 120$
$0 = {x}^{2} - x - 6$
$0 = {x}^{2} + 2 x - 3 x - 6$
$0 = x \left(x + 2\right) - 3 \left(x + 2\right)$
$0 = \left(x + 2\right) \left(x - 3\right)$
$x + 2 = 0 \mathmr{and} x - 3 = 0$
$x = \text{-} 2 \mathmr{and} x = 3$

Since $x$ is the number of hours for the slower car to reach the suburb, a negative solution can be disregarded as extraneous as negative time does not (yet?) exist. $x = 3$

We know that the speed of the slower car is $\frac{120}{x} \text{km/h}$ and that of the faster car is $\frac{120}{x - 1} \text{km/h}$ so plugging in $x$, we get:
Slower car: $\frac{120}{\left(3\right)} = \frac{120}{3} = 40 \text{km/h}$
Faster car: $\frac{120}{\left(3\right) - 1} = \frac{120}{2} = 60 \text{km/h}$